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Variation of Force-time graph for person opening parachute

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  1. Jul 22, 2015 #1
    1. The problem statement, all variables and given/known data
    upload_2015-7-22_22-18-58.png

    2. Relevant equations
    F∝v

    3. The attempt at a solution
    I chose C because I thought that at time T after opening the parachute, the resisting force on the diver will increase. But it is not so since the answer is B. Is it because the diver has reached terminal velocity? Even so, shouldn't the increase in surface area from the parachute increase resisting force?
     
  2. jcsd
  3. Jul 22, 2015 #2
    Perhaps it means to say that the parachute isn't opened until beyond the graph's end? It does say that terminal velocity is reached at t = T.
     
  4. Jul 22, 2015 #3
    Think about what terminal velocity means. What is the acceleration?

    What must the force of air resistance be when the skydiver reaches terminal velocity?
     
  5. Jul 22, 2015 #4

    DEvens

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    Remember that the question says "Beyond the time shown in the graphs, she opens her parachute." So you should not consider the parachute in answering the question. It is only from release to some time after terminal velocity you should consider.

    You are given that force is proportional to velocity. Probably that isn't accurate, but you are given this as a basis.

    So how does force vary with time? Without solving the differential equation (and without being given the actual equation for wind resistance you can't) you do know some things. When the diver first starts to fall she will be moving very slowly. So for the first little while wind resistance will be very small. That means that velocity increases linearly with time for the first little while. But as soon as velocity starts to grow so does wind resistance.

    Terminal velocity means she reaches a constant speed where wind resistance just equals gravity.

    Reaching terminal velocity is an asymptotic thing. There isn't an abrupt corner between increasing speed and non-increasing. Wind resistance builds up, rate of acceleration gradually approaches zero.

    Now looking at these graphs they all have problems. When you are doing a multiple choice kind of quiz you have to choose the best possible answer.
     
  6. Jul 22, 2015 #5

    BvU

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    Are sure about your relevant equation ?

    And the chute opens beyond the time shown on the t-axis, (as axmls already said), so don't even consider that.

    What about the acceleration for small t ? and when t comes closer to T ?
     
  7. Jul 22, 2015 #6
    So, when nearing T the diver almost reaches terminal velocity, that's why there's no more increase in air resistance. But after opening the parachute, why is the force still constant? I don't get that part of the graph.
     
  8. Jul 22, 2015 #7
    It is constant and equal to the weight of diver.
     
  9. Jul 22, 2015 #8

    DEvens

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    [taps desk, looks at student intently]

    Remember that the question says "Beyond the time shown in the graphs, she opens her parachute." So you should not consider the parachute in answering the question. It is only from release to some time after terminal velocity you should consider.

    The time T is NOT the point of opening the parachute. It is the time of reaching terminal velocity.
     
  10. Jul 22, 2015 #9
    So , all the graphs are not very right, is it?
     
  11. Jul 22, 2015 #10

    DEvens

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    Yes, but I think I have given you enough hints. :wink:
     
  12. Jul 22, 2015 #11
    Oh *facepalm*. thanks. That sentence did not register in my head.
     
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